AD is the median of a triangle ABC and AE is perpendicular to BC.
Prove that AB ² = AD² – BC . DE + (BC/2)^2
[If AC >AB]
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Solution:-
D is the midpoint of BC and AE ⊥ BC.
In right angled triangle ABE,
AB² = AE² + BE² ....(1) - Pythagoras Theorem
AD² = AD² + ED² ....(2) - Pythagoras Theorem
From (1) and (2), we get
AB² = AD² - MD² + BE²
⇒ AB² = AD² - DE² + (BD - DE)²
⇒ AB² = AD² - DE² + BD² + DE² - 2BD × DE
⇒ AB² = AD² - 2BD × DE + BD²
⇒ AB² = AD² - 2(BC/2) × DE + (BC/2)² {∵ BD = DC = BC/2}
⇒ AB² = AD² - BC × DE + BC²/4
Hence proved.
Read more on Brainly.in - https://brainly.in/question/770983#readmore
D is the midpoint of BC and AE ⊥ BC.
In right angled triangle ABE,
AB² = AE² + BE² ....(1) - Pythagoras Theorem
AD² = AD² + ED² ....(2) - Pythagoras Theorem
From (1) and (2), we get
AB² = AD² - MD² + BE²
⇒ AB² = AD² - DE² + (BD - DE)²
⇒ AB² = AD² - DE² + BD² + DE² - 2BD × DE
⇒ AB² = AD² - 2BD × DE + BD²
⇒ AB² = AD² - 2(BC/2) × DE + (BC/2)² {∵ BD = DC = BC/2}
⇒ AB² = AD² - BC × DE + BC²/4
Hence proved.
Read more on Brainly.in - https://brainly.in/question/770983#readmore
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